2017 AMC 10B Problems/Problem 25

Problem

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$ , inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$ . What was her score on the sixth test?

$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution 1

Let the sum of the scores of Isabella's first $6$ tests be $S$ . Since the mean of her first $7$ scores is an integer, then $S + 95 \equiv 0 \text{ (mod 7)}$ , or $S \equiv 3 \text{ (mod 7)}$ . Also, $S \equiv 0 \text{ (mod 6)}$ , so by CRT, $S \equiv 24 \text{ (mod 42)}$ . We also know that $91 \cdot 6 \leq S \leq 100 \cdot 6$ , so by inspection, $S = 570$ . However, we also have that the mean of the first $5$ integers must be an integer, so the sum of the first $5$ test scores must be an multiple of $5$ , which implies that the $6$ th test score is $\boxed{\textbf{(E) } 100}$ .

Solution 2 (Cheap Solution)

By inspection, the sequences $91,93,92,96,98,100,95$ and $93,91,92,96,98,100,95$ work, so the answer is $\boxed{\textbf{(E) } 100}$ . Note: A method of finding this "cheap" solution is to create a "mod chart", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.

2017 AMC 10B (Problems • Answer Key • Resources)
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2017 AMC 10B Problems/Problem 25

Contents

Problem

Solution 1

Solution 2 (Cheap Solution)

See Also